Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
vector<vector<int>> field;
queue<pair<int, int>> answ;
int m, n;
void read() {
  cin >> m >> n;
  field.resize(m);
  field.assign(m, vector<int>(n));
  for (int i = 0; i < (m); i++) {
    for (int j = 0; j < (n); j++) {
      cin >> field[i][j];
    }
  }
}
void decrementRow(int row, int val) {
  for (int i = 0; i < (n); i++) {
    field[row][i] -= val;
  }
}
void decrementCol(int col, int value) {
  for (int i = 0; i < (m); i++) {
    field[i][col] -= value;
  }
}
bool isZero() {
  for (int i = 0; i < (m); i++) {
    for (int j = 0; j < (n); j++) {
      if (field[i][j] != 0) {
        return false;
      }
    }
  }
  return true;
}
void incrementRows(vector<int> &minInRow) {
  for (int i = 0; i < minInRow.size(); i++) {
    decrementRow(i, minInRow[i]);
    for (int j = 1; j <= minInRow[i]; j++) {
      answ.push(make_pair(1, i + 1));
    }
  }
}
void incrementCols(vector<int> &minInCol) {
  for (int i = 0; i < minInCol.size(); i++) {
    decrementCol(i, minInCol[i]);
    for (int j = 1; j <= minInCol[i]; j++) {
      answ.push(make_pair(2, i + 1));
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  read();
  vector<int> minInRow(m, INF), minInColumn(n, INF);
  for (int i = 0; i < (m); i++) {
    for (int j = 0; j < (n); j++) {
      if (field[i][j] < minInRow[i]) {
        minInRow[i] = field[i][j];
      }
      if (field[i][j] < minInColumn[j]) {
        minInColumn[j] = field[i][j];
      }
    }
  }
  bool horizontFirst = m < n ? true : false;
  if (horizontFirst) {
    incrementRows(minInRow);
    for (int i = 0; i < (m); i++) {
      for (int j = 0; j < (n); j++) {
        if (field[i][j] < minInColumn[j]) {
          minInColumn[j] = field[i][j];
        }
      }
    }
    incrementCols(minInColumn);
  } else {
    incrementCols(minInColumn);
    for (int i = 0; i < (m); i++) {
      for (int j = 0; j < (n); j++) {
        if (field[i][j] < minInRow[i]) {
          minInRow[i] = field[i][j];
        }
      }
    }
    incrementRows(minInRow);
  }
  if (isZero()) {
    cout << answ.size() << '\n';
    while (!answ.empty()) {
      pair<int, int> t = answ.front();
      answ.pop();
      if (t.first == 1) {
        cout << "row " << t.second << '\n';
      } else {
        cout << "col " << t.second << '\n';
      }
    }
  } else {
    cout << -1;
  }
  return 0;
}